User:Prof McCarthy/kinematics

Kinematics is the branch of classical mechanics that describes the motion of bodies (objects) and systems (groups of objects) without consideration of the forces that cause the motion. The term is the English version of A.M. Ampere's cinématique, which he constructed from the Greek κίνημα, kinema, derived from κινεῖν, kinein, and means to move (or, more literally to stir).

The study of kinematics is often referred to as the geometry of motion. (See analytical dynamics for more detail on usage). The term kinematics also finds use in robotics, biomechanics and animal locomotion. Further, mathematicians have developed the subject of kinematic geometry.

The use of geometric transformations, also called rigid transformations, to describe the movement of components of a mechanical system simplifies the derivation of its equations of motion, and is central to dynamic analysis.

Kinematic analysis finds the range of movement for a given mechanism, and, working in reverse, kinematic synthesis designs a mechanism for a desired range of motion. In addition, kinematics applies algebraic geometry to the study of the mechanical advantage of a mechanical system, or mechanism.

Rigid transformations
The movement of components of a mechanical system is analyzed by attaching a reference frame to each part and determining how the reference frames move relative to each other. If the structural strength of the parts are sufficient then their deformation can be neglected and rigid transformations used to define this relative movement. This brings geometry into the study of mechanical movement.

Geometry is the study of the properties of figures that remain the same while the space is transformed in various ways---more technically, it is the study of invariants under a set of transformations. Perhaps best known is high school Euclidean geometry where planar triangles are studied under congruent transformations, also called isometries or rigid transformations. These transformations displace the triangle in the plane without changing the angle at each vertex or the distances between vertices. Kinematics is often described as applied geometry, where the movement of a mechanical system is described using the rigid transformations of Euclidean geometry.

The coordinates of points in the plane are two dimensional vectors in R2, so rigid transformations are those that preserve the distance, also known as the Pythagorean theorem. The set of rigid transformations in an n-dimensional space is called the special Euclidean group on Rn, and denoted SE(n).

Displacements and motion
The position of one component of a mechanical system relative to another is defined by introducing a reference frame, say M, on one that moves relative to a fixed frame, F, on the other. The rigid transformation, or displacement, of M relative to F defines the relative position of the two components. A displacement consists of the combination of a rotation and a translation.

The set of all displacements of M relative to F is called the configuration space of M. A smooth curve from one position to another in this configuration space is a continuous set of displacements, called the motion of M relative to F. The motion of a body consists of a continuous set of rotations and translations.

Matrix representation
The combination of a rotation and translation in the plane R2 can be represented by 3x3 matrix matrices, known as homogeneous transforms. The 3x3 homogenous transform is constructed from a 2x2 rotation matrix [A(φ)]] and the 2x1 translation vector d=(dx, dy), as
 * $$ [T(\phi, \mathbf{d})] = \begin{bmatrix} A(\phi) & \mathbf{d} \\ 0, 0 & 1\end{bmatrix}

= \begin{bmatrix} \cos\phi & -\sin\phi & d_x \\ \sin\phi & \cos\phi & d_y \\ 0 & 0 & 1\end{bmatrix}.$$ These homogeneous transforms perform rigid transformations on the points in the plane z=1, that is on points with coordinates p=(x, y, 1).

In particular, let p define the coordinates of points in a reference frame M coincident with a fixed frame F, then when the origin of M is displaced by the translation vector d relative to the origin of F and rotated by the angle φ relative to the x-axis of F, then the new coordinates in F of points in M are given by
 * $$ \textbf{P} = [T(\phi, \mathbf{d})]\textbf{p} = \begin{bmatrix} \cos\phi & -\sin\phi & d_x \\ \sin\phi & \cos\phi & d_y \\ 0 & 0 & 1\end{bmatrix}\begin{Bmatrix}x\\y\\1\end{Bmatrix}.$$

Homogeneous transforms represent affine transformations. This formulation is necessary because a translation is not a linear transformation of R2. However, using projective geometry, so that R2 is considered to be a subset of R3, translations become affine linear transformations.

Pure translation
If a rigid body moves so that its reference frame M does not rotate relative to the fixed frame F, the motion is said to be pure translation. In this case, the trajectory of every point in the body is an offset of the trajectory d(t) of the origin of M, that is,
 * $$ \textbf{P}(t)=[T(0,\textbf{d}(t))]\textbf{p} = \textbf{d}(t) + \textbf{p}.$$

Thus, for bodies in pure translation the velocity and acceleration of every point in the body is the same as for the origin,
 * $$ \textbf{V}=\dot{\textbf{P}}(t) = \dot{\textbf{d}}(t),\quad \textbf{A}=\ddot{\textbf{P}}(t) = \ddot{\textbf{d}}(t),$$

where the dot denotes the derivative with respect to time. Recall the coordinate vector p in M is constant, so its derivative is zero.

Trajectories of points in a moving body
Important formulas in the kinematics of a moving body define the velocity and acceleration of points in the body as they trace trajectories in the plane, or three dimensional space. This is particularly important for the center of mass of a body, which is used to derive equations of motion using either Newton's second law or Lagrange's equations.

Position
In order to define these formulas, the movement of a component B of a mechanical system is defined by the set of rotations [A(t)] and translations d(t) assembled into the homogenous transformation [T(t)]=[A(t), d(t)]. Let p be the coordinates of a point P in B measured in the moving frame M, then the trajectory of this point traced in F is given by
 * $$ \textbf{P}(t)=[T(t)]\textbf{p} =

\begin{Bmatrix} \textbf{P} \\ 1\end{Bmatrix}=\begin{bmatrix} A(t) & \textbf{d}(t) \\ 0 & 1\end{bmatrix} \begin{Bmatrix} \textbf{p} \\ 1\end{Bmatrix}.$$ This notation does not distinguish between P = (X, Y, 1), and P = (X, Y), which is hopefully clear in context.

This equation for the trajectory of P can be inverted to compute the coordinate vector p in M as,
 * $$ \textbf{p} = [T(t)]^{-1}\textbf{P}(t) =

\begin{Bmatrix} \textbf{p} \\ 1\end{Bmatrix}=\begin{bmatrix} A(t)^T & -A(t)^T\textbf{d}(t) \\ 0 & 1\end{bmatrix} \begin{Bmatrix} \textbf{P}(t) \\ 1\end{Bmatrix}.$$ This expression uses the fact that the transpose of a rotation matrix is also its inverse, that is
 * $$ [A(t)]^T[A(t)]=I.\!$$

Velocity
The velocity of the point P along its trajectory P(t) is obtained as the time derivative of its position vector,
 * $$ \textbf{V}_P = \frac{d}{dt}\big([T(t)]\textbf{p}\big) =

\begin{Bmatrix} \textbf{V}_P \\ 0\end{Bmatrix} = \begin{bmatrix} \dot{A}(t) & \dot{\textbf{d}}(t) \\ 0 & 0 \end{bmatrix} \begin{Bmatrix} \textbf{p} \\ 1\end{Bmatrix}.$$ The dot denotes the derivative with respect to time, and because p is constant its derivative is zero.

This formula can be modified to obtain the velocity of P by operating on its trajectory P(t). Substitute the inverse transform for p into the velocity equation to obtain
 * $$\textbf{V}_P = \frac{d}{dt}([T(t)])[T(t)]^{-1}\textbf{P}(t) = \begin{Bmatrix} \textbf{V}_P \\ 0\end{Bmatrix} = \begin{bmatrix} \dot{A}A^T & -\dot{A}A^T\textbf{d} + \dot{\textbf{d}} \\ 0 & 0 \end{bmatrix}

\begin{Bmatrix} \textbf{P}(t) \\ 1\end{Bmatrix}=[S]\textbf{P}.$$ The matrix [S] is given by
 * $$ [S] = \begin{bmatrix} \Omega & -\Omega\textbf{d} + \dot{\textbf{d}} \\ 0 & 0 \end{bmatrix}$$

where
 * $$ [\Omega] = \dot{A}A^T,$$

is the angular velocity matrix.

Multiplying by the operator [S], the formula for the velocity VP takes the form
 * $$\textbf{V}_P = [\Omega](\textbf{P}-\textbf{d}) + \dot{\textbf{d}} = \omega\times \textbf{R}_{P/O} + \textbf{V}_O,$$

where the vector ω is the angular velocity vector obtained from the components of the matrix [Ω], the vector
 * $$ \textbf{R}_{P/O}=\textbf{P}-\textbf{d},$$

is the position of P relative to the origin O of the moving frame M, and
 * $$\textbf{V}_O=\dot{\textbf{d}},$$

is the velocity of the origin O.

Acceleration
The acceleration of a point P in a moving body B is obtained as the time derivative of its velocity vector,
 * $$\textbf{A}_P = \frac{d}{dt}\textbf{V}_P = \frac{d}{dt}\big([S]\textbf{P}\big)=[\dot{S}]\textbf{P} + [S]\dot{\textbf{P}} = [\dot{S}]\textbf{P} + [S][S]\textbf{P} .$$

This equation can be expanded by first computing
 * $$ [\dot{S}] = \begin{bmatrix} \dot{\Omega} & -\dot{\Omega}\textbf{d}  -\Omega\dot{\textbf{d}}  + \ddot{\textbf{d}} \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} \dot{\Omega} & -\dot{\Omega}\textbf{d}  -\Omega\textbf{V}_O  + \textbf{A}_O \\ 0 & 0 \end{bmatrix}$$

and
 * $$ [S]^2 = \begin{bmatrix} \Omega & -\Omega\textbf{d} + \textbf{V}_O \\ 0 & 0 \end{bmatrix}^2 = \begin{bmatrix} \Omega^2 & -\Omega^2\textbf{d} + \Omega\textbf{V}_O \\ 0 & 0 \end{bmatrix}.$$

The formula for the acceleration AP can now be obtained as
 * $$ \textbf{A}_P = \dot{\Omega}(\textbf{P} - \textbf{d}) + \textbf{A}_O + \Omega^2(\textbf{P}-\textbf{d}),$$

or
 * $$ \textbf{A}_P = \alpha\times\textbf{R}_{P/O} + \omega\times\omega\times\textbf{R}_{P/O} + \textbf{A}_O,$$

where α is the angular acceleration vector obtained from the derivative of the angular velocity matrix,
 * $$\textbf{R}_{P/O}=\textbf{P}-\textbf{d},$$

is the relative position vector, and
 * $$\textbf{A}_O = \ddot{\textbf{d}}$$

is the acceleration of the origin of the moving frame M.

Coordinates for particle trajectories
The trajectory of a particle P is defined by its coordinate vector P measured in a fixed reference frame F. As the particle moves, its coordinate vector P(t) traces a curve in space, given by
 * $$ \textbf{P}(t) = X(t)\vec{i} + Y(t)\vec{j} + Z(t)\vec{k},$$

where i, j, and k are the unit vectors along the X, Y and Z axes of F, respectively. There are a number of ways to define the functions X(t), Y(t) and Z(t) to match constraints imposed on the trajectory. Here, the particular cases of cylindrical coordinates is presented.

Cylindrical Coordinates
If the particle P moves on the surface of a circular cylinder, it is possible to align the Z axis of the fixed frame F with the axis of the cylinder. Then, the angle θ around this axis in the X-Y plane can be used to define the trajectory as,
 * $$ \textbf{P}(t) = R\cos\theta(t)\vec{i} + R\sin\theta(t)\vec{j} + Z(t)\vec{k}.$$

The cylindrical coordinates for P(t) can be simplified by introducing the radial and tangential unit vectors,
 * $$ \textbf{e}_r = \cos\theta(t)\vec{i} + \sin\theta(t)\vec{j}, \quad \textbf{e}_t = -\sin\theta(t)\vec{i} + \cos\theta(t)\vec{j}.$$

Using this notation, P(t) takes the form,
 * $$ \textbf{P}(t) = R\textbf{e}_r + Z(t)\vec{k},$$

where R is constant.

The velocity of VP is the time derivative of the trajectory P(t),
 * $$ \textbf{V}_P = \frac{d}{dt}(R\textbf{e}_r + Z(t)\vec{k}) = R\dot{\theta}\textbf{e}_t + \dot{Z}(t),$$

where
 * $$ \frac{d}{dt}\textbf{e}_r = \dot{\theta}\textbf{e}_t. $$

If the trajectory P(t) is not constrained to lie on a circular cylinder, then the radius R varies with time, so we have
 * $$ \textbf{P}(t) = R(t)\textbf{e}_r + Z(t)\vec{k},$$

and
 * $$ \textbf{V}_P = \dot{R}\textbf{e}_r + R\dot{\theta}\textbf{e}_t + \dot{Z}(t)\vec{k}.$$

In this case, the acceleration AP, which is the time derivative of the velocity VP, is given by
 * $$ \textbf{A}_P = \frac{d}{dt}(\dot{R}\textbf{e}_r + R\dot{\theta}\textbf{e}_t + \dot{Z}(t)\vec{k}) = (\ddot{R} - R\dot{\theta}^2)\textbf{e}_r + (R\ddot{\theta} + 2\dot{R}\dot{\theta})\textbf{e}_t + \ddot{Z}(t)\vec{k}.$$

Planar circular trajectories
A special case of a particle trajectory on a circular cylinder occurs when there is no movement along the Z axis, in which case
 * $$ \textbf{P}(t) = R(t)\textbf{e}_r + Z_0\vec{k},$$

where R and Z0 are constants. In this case, the velocity VP is given by
 * $$ \textbf{V}_P = \frac{d}{dt}(R\textbf{e}_r + Z_0\vec{k}) = R\dot{\theta}\textbf{e}_t.$$

The acceleration AP of the particle P, is not given by
 * $$ \textbf{A}_P = \frac{d}{dt}(R\dot{\theta}\textbf{e}_t) = - R\dot{\theta}^2\textbf{e}_r + R\ddot{\theta}\textbf{e}_t.$$

The components
 * $$ a_r = - R\dot{\theta}^2, \quad a_t = R\ddot{\theta},$$

are called the radial and tangential components of acceleration, respectively.